Problem: Find $\cos\left(345^\circ\right)$ exactly using an angle addition or subtraction formula. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{-1-\sqrt{3}}{2}$ (Choice B) B $\dfrac{\sqrt{6}-\sqrt{2}}{4}$ (Choice C) C $\dfrac{\sqrt{6}+\sqrt{2}}{4}$ (Choice D) D $\dfrac{\sqrt{3}-1}{2}$
The strategy First, we should rewrite the given angle $345^\circ$ as the sum or difference of two special angles. Then, we can use the cosine addition or subtraction identities in order to evaluate $\cos\left(345^\circ\right)$. [How do we find the trigonometric value of a sum or difference?] Rewriting $345^\circ$ We can rewrite $345^\circ$ as follows. $\begin{aligned}345^\circ&=135^\circ+210^\circ\end{aligned}$ In other words, $345^\circ$ is the sum of the special angles $135^\circ$ and $210^\circ$. Evaluating $\cos\left(345^\circ\right)$ Using the cosine addition identity, we get the following. $\begin{aligned} \cos\left(345^\circ\right)&= \cos\left(135^\circ+210^\circ\right) \\\\\\ &= \cos \left(135^\circ\right) \cos \left(210^\circ\right) - \sin \left(135^\circ\right) \sin \left(210^\circ\right) \\\\\\ &=\left(-\dfrac{\sqrt{2}}{2}\right) \left(-\dfrac{\sqrt{3}}{2}\right) - \left(\dfrac{\sqrt{2}}{2}\right) \left(-\dfrac{1}{2}\right) \\\\\\ &=\left(\dfrac{\sqrt{6}}{4}\right) - \left(-\dfrac{\sqrt{2}}{4}\right)\\\\\\ &=\dfrac{\sqrt{6}+\sqrt{2}}{4} \end{aligned}$ Summary $\cos\left(345^\circ\right) = \dfrac{\sqrt{6}+\sqrt{2}}{4}$